118 Pascal ' s Triangle

Topic:

Given numrows, generate the first numrows of Pascal ' s triangle.

For example, given numrows = 5,
Return

[     1], [    1,2,1], [   1,3,3,1], [  1,4,6,4,1]

Links: http://leetcode.com/problems/pascals-triangle/

A brush, the problem is simple but the solution is not good.

1. The first problem is the upper limit of NumRows and range (N), because range starts with 0, so range (numRows-1)
2. List.reverse () is in place, the reversed (list) return value is a new list
3. cur = cur.extend (second_half) return is None!!! Cur.extend (second_half) is enough, this error is not usually possible at all, too low
4. NumRows = = 1 is returned as a special example

There is a summary that seems to avoid 4), often I will use prev, cur two list, if only one word seems to be able to judge a boundary situation less .

1 classsolution (object):2     defGenerate (Self, numrows):3         ifNumRows = =0:4             return []5         ifNumRows = = 1:6             return[[1]]7result = [[1]]8prev = [1]9Cur = []Ten          One          forRowinchRange (numRows-1): ACur.append (1) -              forIdxinchRange (ROW/2): -Cur.append (Prev[idx] + prev[idx + 1]) theSecond_half =reversed (cur) -             ifRow% 2: -Cur.append (PREV[ROW/2] + PREV[ROW/2 + 1]) - cur.extend (second_half) + result.append (cur) -Prev, cur =cur, [] +         returnResult

118 Pascal ' s Triangle